Solve for $x$ : $7\sqrt{x} + 6 = 3\sqrt{x} + 7$
Explanation: Subtract $3\sqrt{x}$ from both sides: $(7\sqrt{x} + 6) - 3\sqrt{x} = (3\sqrt{x} + 7) - 3\sqrt{x}$ $4\sqrt{x} + 6 = 7$ Subtract $6$ from both sides: $(4\sqrt{x} + 6) - 6 = 7 - 6$ $4\sqrt{x} = 1$ Divide both sides by $4$ $\frac{4\sqrt{x}}{4} = \frac{1}{4}$ Simplify. $\sqrt{x} = \dfrac{1}{4}$ Square both sides. $\sqrt{x} \cdot \sqrt{x} = \dfrac{1}{4} \cdot \dfrac{1}{4}$ $x = \dfrac{1}{16}$